[blazenfs]

A multiple-choice test consists of 60 questions is question has four possible answers of which one is correct it all answers are random guesses estimate the probability of getting at least 20% correct

[TonyChamberlain]

The answer to this question is easily solved using something called The Binomial Probability Distribution. The formula will give you the probability of getting exactly k things (successes) if you try N number of times. P = probability of success, q = 1-p (failure). N = total number of tries, k = number of successes.

Prob = nCk * p^k * q^n-k.

nCk stands for n Choose k. Meaning how many ways can I choose k items from N things. The formula for it is n! / (k!)(n-k!).

In our case p = .25 q=.75 n = 60 and since 12 is 20% of 60 k=12.

P(12) = 60C12 * (.25)^12 * (.75)^48 = 1399358855975 * 5.96e-8 * 1.0068e-6 = 8.3%

However, this only answers the question for getting EXACTLY 20% right. To get at least 20% we need to figure out P(12)+P(13)+P(14) +... P(60). However we can save some time by using the concept that if the probably of something happening is p, 1-p is the probability of it not happening and they both sum to 1. So we can just figure the probability of NOT getting 20% and subtracting that from 1. 1 - (P(11) + P(10) + P(9) + P(8)... +P(1) ).

P(11) = 60c11 * (.25)^11 * (.75)^49 = 6.16%
P(10) = 60c10 * (.25)^10 * (.75)^50 = 4.07%
P(9) = 60c9 * (.25)^9 * (.75)^51 = 2.40%
P(8) = 60c8 * (.25)^8 * (.75)^52 = 1.24%
P(7) = 60c7 * (.25)^7 * (.75)^53 = 0.56%
P(6) = 60c6 * (.25)^6 * (.75)^54 = 0.22%
P(5) = 60c5 * (.25)^5 * (.75)^55 = 0.07%
P(4) = 60c4 * (.25)^4 * (.75)^56 = 0.02%
P(3) = 60c3 * (.25)^3 * (.75)^57 = 0.0004%
P(2) = 60c2 * (.25)^2 * (.75)^58 = 0.00006%
P(1) = 60c1 * (.25)^1 * (.75)^59 = 0.0000006% (almost impossible to randomly get only 1 right!! )

So add all this up then subtract it from 1 = 85.24% Which is your just of getting AT LEAST 20% (12 problems) correct. Which makes sense. If you just randomly got 1 in 4 right you'd get 15.. so you should have a pretty good chance of getting 12.

[ebaines]

Another way to estimate this is to use the fact that binomial probability distribution for N trials is normally distributed, with a mean of Np and a variance of Npq. So the probability of getting at least 12 correct is equivalent to 1- probability of getting less than 12 correct, which is the cumulative distribution function for this distribution at x = 12. This technique gives a good estimate - it's not as exact as the calculation in the previous post, but may be "good enough" given that the question asked for an estimate.

Also, this may be splitting hairs, but to be complete the exact calculation should also include P(0), which is admittedly very small and won't affect the answer much at all.

[TonyChamberlain]

Oops yep.. forgot about P(0). Thanks :)